- 三角函数的诱导公式及应用
- 共6354题
计算:cot260°+tan35°+tan10°cot415°=______.
正确答案
1
解析
解:cot260°+tan35°+tan10°cot415°=cot80°+tan35°+tan10°cot55°
=tan10°+tan35°+tan10°tan35°=tan(10°+35°)(1-tan10°tan35°)+tan10°tan35°
=1×(1-tan10°tan35°)+tan10°tan35°=1,
故答案为:1.
已知cos(75°+θ)=,θ为第三象限角,求cos(255°+θ)+(435°+θ)的值.
正确答案
解:∵θ为第三象限角,不妨假设180°≤θ≤270°,则 255°≤75°+θ≤345°,
故 sin(75°+θ)=-=-
.
cos(-225°-θ)+sin(435°+θ)
=cos(180°+45°+θ)+sin(75°+θ)
=-cos(75°+θ-30°)+sin(75°+θ)
=-[cos(75°+θ)cos30°+sin(75°+θ)sin30°]+sin(75°+θ)
=-(-
)-
=
.
解析
解:∵θ为第三象限角,不妨假设180°≤θ≤270°,则 255°≤75°+θ≤345°,
故 sin(75°+θ)=-=-
.
cos(-225°-θ)+sin(435°+θ)
=cos(180°+45°+θ)+sin(75°+θ)
=-cos(75°+θ-30°)+sin(75°+θ)
=-[cos(75°+θ)cos30°+sin(75°+θ)sin30°]+sin(75°+θ)
=-(-
)-
=
.
sin420°的值( )
正确答案
解析
解:sin420°=sin(360°+60°)=sin60°=.
故选A.
若sin(180°+α)=-,0°<α<90°.求
的值.
正确答案
解:由sin(180°+α)=-,α∈(0°,90°),
得sin α=,cos α=
,
∴原式=
=
=
=2.
解析
解:由sin(180°+α)=-,α∈(0°,90°),
得sin α=,cos α=
,
∴原式=
=
=
=2.
cos40°+cos60°+cos80°+cos160°=______.
正确答案
解析
解:cos40°+cos60°+cos80°+cos160°
=+cos40°+cos80°-cos20°
=+cos(60°-20°)+cos(60°+20°)-cos20°
=+cos60°cos20°+sin60°sin20°+cos60°cos20°-sin60°sin20°-cos20°
=+2cos60°cos20°-cos20°=
+cos20°-cos20°=
.
故答案为:.
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